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A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.10 m/s2 for 14.0 s.
2. Maintain a constant velocity for the next 1.90 min.
3. Apply a constant negative acceleration of −9.06 m/s2 for 3.25 s.
m/s
a) What was the total displacement for the trip?
b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?

Respuesta :

jayilych4real

The total displacement for the trip is 5510 m


Calculation and Parameters

During part 1, the distance travelled is

(1/2) a t^2

= (1/2)(2.77)(225)

= 311.6 m

The final speed gotten is (2.77)(15)

= 41.55 m/s.

The average speed for that interval is half of that or 20.8 m/s

During part 2, the distance traveled is

41.55 m/s*2.05 min*60 sec/min

= 5110 m/s

During part 3, the speed drops from 41.55 m/s to

41.55 - (9.47)(4.39)

= 0

The average speed during this interval is

41.55/2 - 20.78 m/s

The distance travelled is 20.78*4.39

= 91.2 m/s

Total distance travelled = 312 + 5110 + 91 = 5513 m

The average speeds for legs 1,2,and 3 of the trip as well as for the complete trip are:

  1. 20.8 m/s,
  2. 41.6 m/s,
  3. 20.8 m/s,
  4. 38.7 m/s

Read more about average speed here:

https://brainly.com/question/6504879

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